Cesium chloride crystallizes as cubic lattice and has a density of `4.0 g cm^(-3)` . Calculate the length of the edge of the unit cell of cesium chloride. (Molar of CsCl = `168.5 g mol^(-1))`

Explanation

To calculate the length of the edge of the unit cell of cesium chloride (CsCl), we can follow these steps:

### Step 1: Understand the formula for density
The density (d) of a substance is given by the formula:
\[ d = \frac{Z \cdot M}{A^3 \cdot N_0} \]
where:
- \( Z \) = number of formula units (or molecules) per unit cell
- \( M \) = molar mass of the substance (in grams per mole)
- \( A \) = edge length of the unit cell (in cm)
- \( N_0 \) = Avogadro's number (\( 6.022 \times 10^{23} \) mol\(^{-1}\))

### Step 2: Identify the values
From the question, we have:
- Density of CsCl, \( d = 4.0 \, \text{g/cm}^3 \)
- Molar mass of CsCl, \( M = 168.5 \, \text{g/mol} \)

For cesium chloride (CsCl), it crystallizes in a simple cubic lattice, which means:
- \( Z = 1 \) (since there is one CsCl formula unit per unit cell)

### Step 3: Substitute the values into the density formula
We can rearrange the density formula to solve for \( A^3 \):
\[ A^3 = \frac{Z \cdot M}{d \cdot N_0} \]

Substituting the known values:
\[ A^3 = \frac{1 \cdot 168.5 \, \text{g/mol}}{4.0 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 4: Calculate \( A^3 \)
Calculating the denominator:
\[ 4.0 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} = 2.4088 \times 10^{24} \, \text{g/cm}^3 \]

Now, substituting back:
\[ A^3 = \frac{168.5}{2.4088 \times 10^{24}} \]

Calculating \( A^3 \):
\[ A^3 \approx 6.99 \times 10^{-23} \, \text{cm}^3 \]

### Step 5: Calculate \( A \)
To find \( A \), take the cube root:
\[ A = (6.99 \times 10^{-23})^{1/3} \]

Calculating \( A \):
\[ A \approx 4.12 \times 10^{-8} \, \text{cm} \]

### Step 6: Convert to picometers
To convert centimeters to picometers (1 cm = \( 10^{10} \) pm):
\[ A \approx 4.12 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{pm/cm} \]
\[ A \approx 412 \, \text{pm} \]

### Final Answer
The length of the edge of the unit cell of cesium chloride is approximately **412 picometers**.

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